Question: You have found the following ages (in years) of all 4 seals at your local zoo: $ 7,\enspace 1,\enspace 8,\enspace 15$ What is the average age of the seals at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 4 seals at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{7 + 1 + 8 + 15}{{4}} = {7.8\text{ years old}} $ Find the squared deviations from the mean for each seal. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $7$ years $-0.8$ years $0.64$ years $^2$ $1$ year $-6.8$ years $46.24$ years $^2$ $8$ years $0.2$ years $0.04$ years $^2$ $15$ years $7.2$ years $51.84$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.64} + {46.24} + {0.04} + {51.84}} {{4}} $ $ {\sigma^2} = \dfrac{{98.76}}{{4}} = {24.69\text{ years}^2} $ The average seal at the zoo is 7.8 years old. The population variance is 24.69 years $^2$.